# Math Help - trigonometric problems

1. ## trigonometric problems

I am on the verge of an A in math, its finals week, and doing these extra credits will help me immensely. There are 10 problems, I figured out 7 of them, and I cant get these for the life of me. Trig is my weak point, and I really just dont understand the last one at all. I would love some help

1) Verify the identity analytically

(SinX/cosXtanX)+[(cotX-tanX)/(cotX+tanX)]= 1 + cos2X

2) Solve analytically on the interval [2pi, 4pi)

Sin2x Cot2x = sin2x

4) An aircraft carrier is traveling on a steady course with a bearing of 30 degrees at 32 mph. Planes on the carrier have enough fuel for 2.6 hours of flight when traveling at a speed of 520 mph. One of the planes takes off at a bearing of 338 degrees, and then turns and heads in a straight line, so as to be able to catch the carrier and land at the exact instant that its fuel runs out. If the pilot left at 2:00pm, at what time did he turn to head to the carrier?

2. #2:

The plane can fly 1352 miles if it can fly at 520 mph for 2.6 hrs.

In that time, the carrier travels 30*2.6=83.2 miles.

Let OA=x and AB=y. Then x+y=1352.

By the law of cosines:

$y=\sqrt{x^{2}+(83.2)^{2}-2x(83.2)cos(52)}$

Therefore, the total distance is:

$x+\sqrt{x^{2}+(83.2)^{2}-2x(83.2)cos(52)}=1352$

Solve for x and y will follow.

What you need is the distance travelled by the plane from OA=x.

Once you find this, divide it by the planes rate and add it to 2:00 pm

I have included a diagram. It is not too accurate, but I hope it shows the idea.

3. Hello, nauticaricky! Let me give you some hints:

Originally Posted by nauticaricky
1) Verify the identity analytically

(SinX/cosXtanX)+[(cotX-tanX)/(cotX+tanX)]= 1 + cos2X
Converting everything to $\sin x\ \text{ and }\cos x$ will make this easier. Make sure you are comfortable with simplifying fractions, and you will need to know these trig identities:

$\cos 2x = \cos^2 x - \sin^2 x$

$\sin^2 x + \cos^2 x = 1$

Originally Posted by nauticaricky
2) Solve analytically on the interval [2pi, 4pi)

Sin2x Cot2x = sin2x
Subtract both sides by $\sin 2x$, factor, then solve.

Originally Posted by nauticaricky
4) An aircraft carrier is traveling on a steady course with a bearing of 30 degrees at 32 mph. Planes on the carrier have enough fuel for 2.6 hours of flight when traveling at a speed of 520 mph. One of the planes takes off at a bearing of 338 degrees, and then turns and heads in a straight line, so as to be able to catch the carrier and land at the exact instant that its fuel runs out. If the pilot left at 2:00pm, at what time did he turn to head to the carrier?
I suggest drawing a diagram. Draw a line representing the distance the carrier travels (the length of which will be its speed of 32 mph times the amount of time it travels during the plane's trip), draw another line from the starting point at the appropriate angle representing the distance the plane travels before turning, and then connect the end of that line to the end of the carrier's path to represent the distance taken to get back. Label everything, and knowing that the duration of the plane's flight is 2.6 hours, you should be able to find the distance that the carrier and the plane travel. Then, use the fact that the plane travels for $\text{speed}\cdot\text{time} = 520t_0$ miles before turning and then an additional $520t_1$ miles before making it back to the carrier. You should then be able to use the Law of Cosines, $c^2 = a^2 + b^2 - 2ab\cos C,\ \text{where }C\text{ is the angle opposite side }c$, to solve for $t_0$, which is the time that elapses before the plane turns.

I think that should work, but I haven't carried the problem all the way through, so if you run into difficulties let me know.

4. 1) Verify the identity analytically

(SinX/cosXtanX)+[(cotX-tanX)/(cotX+tanX)]= 1 + cos2X

can you help more on this one? I still cant get it!

5. Originally Posted by nauticaricky
1) Verify the identity analytically

(SinX/cosXtanX)+[(cotX-tanX)/(cotX+tanX)]= 1 + cos2X

can you help more on this one? I still cant get it!
The first term on the left is obviously 1, so all we need to do is look at the second term.
$\frac{cot(x) - tan(x)}{cot(x) + tan(x)}$

$= \frac{\frac{cos(x)}{sin(x)} - \frac{sin(x)}{cos(x)}}{\frac{cos(x)}{sin(x)} + \frac{sin(x)}{cos(x)}}$

$= \frac{\frac{cos(x)}{sin(x)} - \frac{sin(x)}{cos(x)}}{\frac{cos(x)}{sin(x)} + \frac{sin(x)}{cos(x)}} \cdot \frac{sin(x)~cos(x)}{sin(x)~cos(x)}$

$= \frac{cos^2(x) - sin^2(x)}{cos^2(x) + sin^2(x)}$

$= \frac{cos(2x)}{1} = cos(2x)$

-Dan